MTH 847: PDE I (Fall 2017) Exam 2, 2017.12.8 Name: Standard exam rules apply: You are not allowed to give or receive help from other students. All electronic devices must be turned off for the duration of the test and stowed. This includes phones, pagers, laptops, tablets, e-readers, and calculators. The only things allowed on your desk are: Your writing implements, including also corrector fluids or erasers or similar. A water bottle or other drink. This booklet. This exam lasts from 11:30 12:20. Students may not leave the room until after 12:00; students may not (re)enter the room after 12:00. INSTRUCTOR USE ONLY: Q# pts MAX 1A 4 1B 4 2A 2 2B 2 3A 4 3B 4 4 4 TOTAL 24 1
Q1. Let φ C 2 (R R 3 ) be a solution to the linear wave equation φ = 0. Let λ R + and denote by E the set E := {(t,x) R R 3 t λ, x 1}. A. (4pts) Suppose λ < 1. Show that there exists a solution φ to the wave equation, that is not identically zero, but vanishes on the set E. B. (4pts) Suppose λ > 1. Show that any solution φ that vanishes on the set E must be identically zero. (Hint: strong Huygens principle [alternatively the Kirchhoff formula] can be useful.) 2
Solutions. A. Let ɛ < 1 λ. Take any g Cc (B(0,ɛ)) and solve the wave equation with initial data φ(0,x) = g(x) and t φ(0,x) = 0. By the finite speed of propagation property we have that φ 0 whenever x t + ɛ. Our initial choice of ɛ then guarantees that E {φ = 0}. If g is chosen to be nontrivial, the solution is also nontrivial. B. Consider the points (t,x) with t = 1 and x < 1. Denote by g(x) = φ( 1,x) and h(x) = t φ( 1,x). By the Kirchoff formula we have that φ(1,x) = 1 g(x + 2ω) + 2 4π ν g(x + 2ω) + 2h(x + 2ω) dω. S 2 Notice that when x < 1, we have that x+2ω > 1 by triangle inequality, and hence ( 1,x+2ω) E. This means that the integrand vanishes by assumption. Therefore φ(1, x) 0 for all x. Similarly we conclude that t φ(1,x) 0 for all x, and hence φ is a solution to the linear wave equation such that φ(1,x) = t φ(1,x) 0 and by uniqueness of solutions we have that φ 0. 3
Q2. Suppose φ C 2 (R R) solves 2 ttφ + 2 xxφ = 0 φ(0,x) = 0 t φ(0,x) = h(x) where the data h C 2 (R) is such that h(x) = 0 for all x 1; and h(x) > 0 for all x < 1. A. (2pts) Find all space-time points (t,x) where φ(t,x) = 0. B. (2pts) Find all space-time points (t,x) where t φ(t,x) = 0. 4
Solutions. The answer to both parts follow from applying D Alembert s formula φ(t,x) = 1 x+t 2 [g(x + t) + g(x t) + h(y) dy]. A. In this part g(x) = 0. So when x t + 1 we have that h(y) vanishes in the integrand, and hence φ(t,x) = 0 there. On the other hand, when x < t + 1, φ(t,x) is equal to the integral of a non-trivial, non-negative continuous function, and hence is positive. B. Setting ψ = t φ, we have that ψ solves the wave equation with data ψ(0,x) = h(x) and t ψ(0,x) = 0. So applying D Alembert s formula we have that x t t φ(t,x) = 1 [h(x + t) + h(x t)]. 2 Hence t φ(t,x) > 0 whenever (t,x) { x + t < 1} { x t < 1}, and vanishes exactly on its complement. 5
Q3. Consider the Cauchy problem where (t,x) R R, and Σ = {t = x }. t φ + x x φ = 0 φ Σ = f A. (4pts) Give an example of a function f C 1 (R R) such that there does not exist a C 1 (R R) solution to the Cauchy problem. Justify your example. B. (4pts) Give an example of a function f C 1 (R R) such that there exists infinitely many C 1 (R R) solutions to the Cauchy problem. Justify your example. 6
Solutions. A. The corresponding integral curves are x = Ce t for C R. Hence there are multiple ways to answer this question. 1. We can use the fact that Σ is not smooth: if we let f (t,x) = t, then if φ is any C 1 function agreeing with f along Σ, we must have t φ(0,0) + x φ(0,0) = t φ(0,0) x φ(0,0) = 1. This means that the equation t φ + x x φ cannot hold at the origin. 2. We can use the fact that the integral curves intersect Σ more than once: Let C = e 2 for example, then the system x = e t 2 and x = t has two distinct solutions with different values of t. So setting f (t,x) = t again means that we cannot have a solution, since the PDE implies φ must be constant along the integral curves. 3. We can use the fact that the integral curve becomes tangent to Σ when C = ±e 1. Here (t,x) = (1,±1). Setting again f (t,x) = t would require that at (t,x) = (1,±1) that t φ(1,±1) + x x φ(1,±1) = 1, ruling out the existence of a solution. B. Let ψ C c (R). Then we can check that setting φ(t,x) = ψ(t ln x ) we have t φ + x x φ = ψ (t ln x ) (1 x 1 x ) = 0. Furthermore, since ψ has compact support we have that φ is also smooth (especially near x = 0). Next observe that if t = x, then t ln x = t lnt > 0. So in particular if ψ and ψ agree on the positive real axis, then the corresponding φ and φ agree on the set Σ. So letting f = φ = ψ(t ln x ) for any ψ described as above, we see that there exists infinitely many solutions to the Cauchy problem. 7
Q4. (4pts) Let Ω R d be a bounded open set with C 1 boundary. Consider the system φ = 0 on (0,T ) Ω with the nonlinear initial-boundary conditions φ(0,x) = t φ(0,x) = 0, x Ω; ( t φ) 3 + e φ n φ = 0, along [0,T ] Ω. Here n φ denotes the outward normal derivative on Ω. Prove that if φ C 2 ([0,T ] Ω;R) solves the system above, then φ 0 on [0,T ] Ω. 8
Solutions. The energy method gives, for every τ (0,T ], 0 = ν ( t) J ds + ν ( t) J ds + ν ( t) J ds = E τ + E 0 + F. {τ} Ω {0} Ω [0,τ] Ω Our choice of initial data means that E 0, the integral over {0} Ω, vanishes. We know that along {τ} Ω, the outward normal ν = t and so E τ 0. We can compute on [0,τ] Ω ( t ) J ν = n m Q t = n φ t φ 0 since the boundary assumption implies that t φ and n φ have opposite signs. Together this implies that E τ = F = 0 for any τ. This means that φ 0 on [0,T ] Ω and hence φ 0. 9